The function f: x → mx/x + n, where m and n are constants, is such that f(-1) = 1 1/2 and f(2) = 9. (a) State the domain of f(x). (b) Find the value of m and n. (c) Evaluate f-1(4).

The function $f(x) = \frac{m}{x} + n$ involves a term with $x$ in the denominator. For the function to be defined, the denominator cannot be zero.

$$x \neq 0$$

The domain of $f(x)$ is all real numbers except 0.
Domain: $\boxed{x \in \mathbb{R}, x \neq 0}$

Given $f(-1) = \frac{3}{2}$:

$$\frac{m}{-1} + n = \frac{3}{2}$$ $$-m + n = \frac{3}{2} \quad (1)$$

Given $f(2) = 9$:

$$\frac{m}{2} + n = 9 \quad (2)$$

Subtract equation (1) from equation (2):

$$\left(\frac{m}{2} + n\right) - (-m + n) = 9 - \frac{3}{2}$$ $$\frac{m}{2} + n + m - n = \frac{18}{2} - \frac{3}{2}$$ $$\frac{3m}{2} = \frac{15}{2}$$

Multiply both sides by 2:

$$3m = 15$$ $$m = 5$$

Using equation (1):

$$-5 + n = \frac{3}{2}$$ $$n = \frac{3}{2} + 5$$ $$n = \frac{3}{2} + \frac{10}{2}$$ $$n = \frac{13}{2}$$

The values are $m=5$ and $n=\frac{13}{2}$.
So, the function is $f(x) = \frac{5}{x} + \frac{13}{2}$.
Values: $\boxed{m=5, n=\frac{13}{2}}$

To find $f^{-1}(4)$, we need to find the value of $x$ such that $f(x) = 4$.
Using the function $f(x) = \frac{5}{x} + \frac{13}{2}$:

$$4 = \frac{5}{x} + \frac{13}{2}$$

Subtract $\frac{13}{2}$ from both sides:

$$4 - \frac{13}{2} = \frac{5}{x}$$ $$\frac{8}{2} - \frac{13}{2} = \frac{5}{x}$$ $$-\frac{5}{2} = \frac{5}{x}$$

Multiply both sides by $2x$:

$$-5x = 5 \times 2$$ $$-5x = 10$$ $$x = \frac{10}{-5}$$ $$x = -2$$

Therefore, $f^{-1}(4) = -2$.
Value: $\boxed{-2}$